Question

The time required (in hours) to reduce 3 mol of \( Fe^{3+} \) ions to \( Fe^{2+} \) ions with 2.0 amperes of current is (1 F = 96500 C mol\(^{-1} \))

• A. 30.2
• B. 40.2
• C. 10.2
• D. 15.2

Hint:

Use Faraday's laws of electrolysis. First, determine the number of moles of electrons required based on the stoichiometry of the reduction reaction. Then, calculate the total charge using \( Q = nF \). Finally, use the relationship \( Q = It \) to find the time, and convert the units to hours if needed.

Answer 1

The Correct Option is B

The reduction of \( Fe^{3+} \) to \( Fe^{2+} \) involves the gain of one electron: $$ Fe^{3+} + e^- \rightarrow Fe^{2+} $$ To reduce 1 mole of \( Fe^{3+} \) ions, 1 mole of electrons is required.
Therefore, to reduce 3 moles of \( Fe^{3+} \) ions, 3 moles of electrons are required.
The total charge (Q) required can be calculated using Faraday's law: $$ Q = nF $$ where \( n \) is the number of moles of electrons and \( F \) is the Faraday constant.
Here, \( n = 3 \) moles of electrons and \( F = 96500 \) C mol\(^{-1} \).
$$ Q = 3 \text{ mol} \times 96500 \text{ C mol}^{-1} = 289500 \text{ C} $$ The current (I) is given as 2.
0 amperes (A), which means 2.
0 Coulombs per second (C s\(^{-1} \)).
The time (t) required to pass this charge can be calculated using the formula: $$ Q = It $$ $$ t = \frac{Q}{I} = \frac{289500 \text{ C}}{2.
0 \text{ C s}^{-1}} = 144750 \text{ s} $$ The question asks for the time in hours.
To convert seconds to hours, we divide by 3600 (since 1 hour = 3600 seconds): $$ t (\text{hours}) = \frac{144750 \text{ s}}{3600 \text{ s hour}^{-1}} = 40.
2083 \text{ hours} $$ Rounding to one decimal place, the time required is 40.
2 hours.